HDU-2016 Multi-University Training Contest 3-Sqrt Bo-大数开方

Sqrt Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0


Problem Description

Let's define the function f(n)=⌊n−−√⌋.

Bo wanted to know the minimum number y which
satisfies fy(n)=1.

note:f1(n)=f(n),fy(n)=f(fy−1(n))

It is a pity that Bo can only use 1 unit of time to calculate this function each time.

And Bo is impatient, he cannot stand waiting for longer than 5 units of time.

So Bo wants to know if he can solve this problem in 5 units of time.

 

Input

This problem has multi test cases(no more than 120).

Each test case contains a non-negative integer n(n<10100).

 

Output

For each test case print a integer - the answer y or
a string "TAT" - Bo can't solve this problem.

 

Sample Input

233
233333333333333333333333333333333333333333333333333333333

 

Sample Output

3
TAT

 

题目意思：

对一个数开方，能在五次之内取整得到0的就输出次数；不能或者是0的话就输出TAT。。

解题思路：

这题longlong暴力就能过，队友很快就过了~然而我当时没想到，只想到了大数……QwQ

吭哧吭哧改了半天……才过…╮(╯▽╰)╭

longlong的代码很简单我就不写了，下面给出我写的大数的…这个模板说不定以后用的上呢~2333

#include <iostream>
using namespace std;
#include <stdlib.h>
#include <string.h>
#define MAXN 1000000
int j,l,size,num,x[MAXN];
char ch[MAXN],ans[MAXN];
void sqrt(char *str)//大数求平方
{
double i,r,n;
size=strlen(str);
if (size%2==1)
{
n=str[0]-48;
l=-1;
}
else
{
n=(str[0]-48)*10+str[1]-48;
l=0;
}
r=0,num=0;
while (true)
{
i=0;
while (i*(i+20*r)<=n)
++i;
--i;
n-=i*(i+20*r);
r=r*10+i;
x[num]=(int)i;
++num;
l+=2;
if (l>=size)
break;
n=n*100+(double)(str[l]-48)*10+(double)(str[l+1]-48);
}

}
int main()
{
while (cin>>ch)
{
int flag=0;
sqrt(ch);
if((x[num-1]==1)&&num==1)//注意判断条件
{
cout<<"1"<<endl;
flag=1;
goto A;
}
for(int i=2; i<=5; ++i)//最多计算五次
{
for (j=0; j<num; ++j)
ans[j]='0'+x[j];//int转char
ans[num]='\0';//注意末尾别忘了加结束符
//cout<<ans<<endl;
sqrt(ans);//开方
if((x[num-1]==1)&&num==1)
{
cout<<i<<endl;//输出次数
flag=1;//标记
break;
}
}
A: if(!flag)
cout<<"TAT"<<endl;//超过五次或者为0时
}
return 0;
}